Download - Lectures on Local Fields
-
7/26/2019 Lectures on Local Fields
1/57
Lectures on Complete Discrete Valuation Fields
1: Discrete Valuation Fields
(1.1). Valuations. One can generalize the properties of the -adic valuation and proceed
to the concept of valuation. Let be an additively written totally ordered abelian group. Add
to a formal element + with the properties + , + + , + (+ ) = + ,
(+ ) + (+
) = +
, for each ; denote = + .
A map : with the properties
(
) = +
= 0
(
) =
(
) +
(
)
( + ) min( ( ) ( ))
is said to be a valuationon ; in this case is said to be a valuation field. The map induces
a homomorphism of to and its value group ( ) is a totally ordered subgroup of .
If ( ) = 0 , then is called thetrivial valuation. It is easy to show that ( 1) = 0, and if
(
)
(
), then
(
)
min(
(
+
)
(
))
min(
(
)
(
)) =
(
);
thus, if ( ) = ( ) then ( + ) = min( ( ) ( )).
(1.2). Basic Objects. Let
=
: (
)
0
,
=
: (
)
0
. Then
coincides with the set of non-invertible elements of . Therefore, is a local ring with the
unique maximal ideal
;
is called the ring of integers (with respect to ), and the field
= is called the residue field, or residue class field. The image of an element
in is denoted by , it is called the residueof in . The set of invertible elements of
is a multiplicative group =
, it is called thegroup of units.
Assume thatchar( ) = char(
). Then char( ) = 0 andchar(
) = 0.
Proof. Suppose that char( ) = = 0. Then = 0 in and therefore in . Hence = char(
).
(1.3).
1. A valuation on issaid tobe discrete if the totally ordered group ( ) is isomorphic
to the naturally ordered group .
For a prime and a non-zero integer let = ( ) be the maximal integer such that
divides . Extend to rational numbers putting ( ) = ( ) ( ); (0) = + .The -adic valuation is a discrete valuation with the ring of integers
=
: is relatively prime to
1
-
7/26/2019 Lectures on Local Fields
2/57
2
The residue field
is a finite field of order
.
2. Let = ( ). For an irreducible polynomial ( ) define ( ) similarly to above.
The map (
) is a discrete valuation with the ring of integers
( )
=
(
) (
)
:
( )
( )
[
]
( )is relatively prime to ( )
and the residue field is [ ] ( ) [ ] which is a finite algebraic extension of .
The field has another discrete valuation trivial on : (
) = deg(
), its residue field
is .
(1.4). Discrete valuations. A field is said to be a discrete valuation field if it admits a
nontrivial discrete valuation (see Example 1 in (1.3)). An element is said to be
a prime element(uniformizing element) if (
) generates the group
(
). Without loss of
generality we shall often assume that the homomorphism : issurjective.
Let be a discrete valuation field, and be a prime element. Then the ring of
integers
is a principal ideal ring, and every proper ideal of
can be written as
for
some 0. In particular,
= . The intersection of all proper ideals of is the zero
ideal.
Each element can be uniquely written as for some and .
Proof. Let be a proper ideal of . Then there exists = min ( ) : and hence
for some unit . It follows that and = . If belongs to
the intersection of all proper ideals in , then ( ) = + , i.e., = 0.
Let = ( ). Then and = for . If 1 = 2, then
+
(
1) = + ( 2). As 1 2 , we deduce = , 1 = 2.
(1.5). Completion. Completion of a discrete valuation field is an object which is easier to
work with than with the original field.
Let be a field with a discrete valuation (as usual, ( ) = ). is a metric space
with respect to the norm = (1 2) (
). So one can introduce the notion of a fundamental
sequence: a sequence (
)
0 of elements of is called a fundamental sequence if for every
real there is ( ) 0 such that (
) for ( ).
If (
) is a fundamental sequence then for every integer there is such that for all
we have (
) . We can assume 1 2
. If for every there
is
such that (
) = (
+1), then (
) and (
) for
, and
hence lim (
) = + . Otherwise lim (
) is finite.
The set of all fundamental sequences forms a ring with respect to componentwise
addition and multiplication. The set of all fundamental sequences (
)
0 with
0 as
+
forms a maximal ideal
of
. The field
is a discrete valuation field with its
discrete valuation defined by ((
)) = lim (
) for a fundamental sequence (
)
0.
Proof. A sketch of the proof is as follows. It suffices to show that is a maximal ideal of
. Let (
) 0 be a fundamental sequence with
0 as
+
. Hence, there is an 0 0 such that
= 0 for 0. Put
= 0 for 0 and
= 1
for 0 .
Then (
)
0 is a fundamental sequence and (
)(
) (1) +
. Therefore, is maximal.
-
7/26/2019 Lectures on Local Fields
3/57
3
(1.6). A discrete valuation field is called a complete discrete valuation field if every
fundamental sequence (
)
0 is convergent, i.e., there exists = lim
with respect to
. A field with a discrete valuation is called acompletionof if it is complete, = ,
and
is a dense subfield in
with respect to
.
Every discrete valuation field has a completion which is unique up to an iso-
morphism over .
Proof. We verify that the field with the valuation is a completion of . is
embedded in by the formula ( ) mod . For a fundamental sequence (
)
0
and real , let 0 0 be such that (
) for all 0. Hence, for
0
we have ((
0)
(
)
0) , which shows that is dense in . Let (( (
)
)
)
be a fundamental sequence in with respect to . Let (0), (1),
be an increasing
sequence of integers such that ( ( )
2
(
)
1)
for 1 , 2 ( ). Then ( (
)
(
))
is a
fundamental sequence in and it is the limit of (( ( )
)
)
with respect to in . Thus,
we obtain the existence of the completion , .
If there are two completions
1 , 1 and
2 , 2 , then we put
( ) = for andextend this homomorphism by continuity from , as a dense subfield in 1, to 1 . It is easy
to verify that the extension
: 1
2 is an isomorphism and 2
= 1 .
We shall denote the completion of the field with respect to by or simply .
1.7. Examples of complete valuation fields. 1. The completion of with respect to of
(1.3) is denoted by and is called thefield of -adic numbers. Certainly, the completion of
with respect to the absolute value of (1.1) is . Embeddings of in for all prime
and in is a tool to solve various problems over . An example is theMinkowskiHasse
Theorem: an equation = 0 for has a nontrivial solution in if and only
if it admits a nontrivial solution in and in for all prime
.
The ring of integers of
is denoted by
and is called the ring of
-adic integers. Theresidue field of is the finite field consisting of elements.
2. The completion of ( ) with respect to
is the formal power series field (( ))
of all formal series +
with
and
= 0 for almost all negative . The
ring of integers with respect to
is [[
]], that is, the set of all formal series
+
0
,
. Its residue field may be identified with .
(1.8). Representatives. For simplicity, we will often omit the index in notations , ,
, . We fix a prime element of .
A set is said to be a set of representatives for a valuation field if , 0
and is mapped bijectively on under the canonical map = . Denote by
rep: the inverse bijective map. For a set denote by ( )+
the set of all sequences
( )
, . Let ( )+
denote the union of increasing sets ( )+
where .
The additive group has a natural filtration
+1
The factor filtration of this filtration is easy to calculate:
+1 .
-
7/26/2019 Lectures on Local Fields
4/57
4
Let be a complete field with respect to a discrete valuation . Let
for each
be an element of
with
(
) =
. Then the map
Rep: (
)
+
(
)
+
rep(
)
is a bijection. Moreover, if ( )
= (0)
then
(Rep(
)) = min
: = 0
.
Proof. The map Rep is well defined, because for almost all
0 we get rep( ) = 0 and the
series rep( ) converges in . If ( ) = ( ) and
= min
: =
then (
) = . Since (
) for
, we deduce that
(Rep(
)
Rep(
)) =
Therefore Rep is injective.
In particular, (Rep( )) = min
: = 0 . Further, let . Then = with
, . We also get =
for some . Let
be the image of in ;
then
= 0 and 1 = rep(
)
+1 . Continuing in this way for 1, we obtain a
convergent series = rep( ) . Therefore, Rep is surjective.
We oftentake
= . Therefore, by the preceding Lemma, every element
can be uniquely expanded as
=
+
and
= 0 for almost all
0
If , we write mod .
(1.9). Units. The group1 + is called the group of principal units 1 and its elements are
called principal units. Introduce alsohigher groups of units:
= 1 +
for
1.
The multiplicative group has a natural filtration 1 2
. We
describe the factor filtration of the introduced filtration on .
Let be a discrete valuation field. Then
(1) The choice of a prime element ( 1 ) induces an isomorphism
.
(2) The canonical map = induces the surjective homomorphism
0: ;
0 maps 1 isomorphically onto .
(3) The map
:
1 +
for induces the isomorphism
of
+1 onto for
1.
-
7/26/2019 Lectures on Local Fields
5/57
5
Proof. (2) The kernel of
0 coincides with 1 and
0 is surjective. (3) The induced map
+1 is a homomorphism, since
(1 + 1
)(1 + 2
) = 1 + ( 1+ 2)
+ 1 2 2
This homomorphism is bijective, since
(1 + rep( )
) = .
Let be not divisible by char(
). Raising to the
th power induces an auto-
morphism of
+1 for
1. If is complete, then the group
for
1 is uniquely
-divisible.
Proof. If = 1 +
with , then 1 +
mod +1 . Absence of nontrivial
-torsion in the additive group implies the first property. It also shows that has no
nontrivial -torsion.
For an element = 1 +
with we have = (1 + 1
) 1 with 1 +1 .
Applying the same argument to 1 and so on, we get an th root of in in the case of
complete .
(1.10). Raising to
th power. Let char( ) =
0. Lemma (1.2) shows that eitherchar(
) = or char( ) = 0. We shall study the operation of raising to the th power. Denote
this homomorphism by :
The first and simplest case is char( ) =
.
Let char( ) = char(
) =
0. Then the homomorphism
maps
injectively into for
1. For
1
(1 +
)
1 +
mod +1
Proof. Since (1 +
)
= 1 +
and there is no nontrivial
-torsion in and , the
assertion follows.
Let be a field of characteristic
0 and let
be perfect, i.e
=
. Then maps the quotient group
+1 isomorphically onto the quotient group +1 for
1.
We now consider the case of char( ) = 0, char( ) = 0. As = 0 in the residue field
, we conclude that and, therefore, for the surjective discrete valuation of we get
( ) = 1.
The number = ( ) = ( ) is called the absolute ramification index of .
Let be a prime element in
. Let
be a set of representatives, and let
0 be the
element of uniquely determined by the relation
0 +1
.
Let be a discrete valuation field of characteristic zero with residue field of
positive characteristic
. Then the homomorphism
maps to for
(
1), and
to +
for
( 1). Moreover, for
(1 +
)
1 +
mod
+1
if
(
1)
(1)
-
7/26/2019 Lectures on Local Fields
6/57
6
(1 +
)
1 + (
+
0 )
mod
+1
if
= (
1)
(2)
(1 +
)
1 +
0 +
mod +
+1 if
( 1) (3)
The induced homomorphisms on the quotient filtration are injective in cases (1), (3) andsurjective in case(3).
If a primitive
th root of unity is contained in , then (1 ) = ( 1) and the
kernel of the induced homomorphisms in case (2)is of order .
If is complete, then
+ =
for
( 1). If is complete and ( 1) ,
then the homomorphism in(2)is injective iff there is no nontrivial -torsion in .
Proof. Let 1 + . We get
(1 + )
= 1 + +
( 1)
2
2 + +
1 +
and ( ) = +
, ( 1)
2
2 =
+ 2
, (
1) = + (
1)
, (
) =
, so
((1 +
)
1) =
(
+ )
if (
)
= (
)
((1 +
)
1)
(
+ )
otherwise
Note (
) ( ) if and only if
( 1). For a unit we obtain the first statement of
the proposition.
Further, the homomorphism
is an isomorphism in case (3) and injective in case (1).
Assume that . From the previous (1 ) = (
1) and (
1)
.
Therefore, the homomorphism
+
0 is not injective. Its kernel 1
0 in this
case is of order
.
If is complete, then due to surjectivity of the homomorphisms in case (3) for
( 1) we get = +1
= +2
= =
. Now let ( 1) be an integer.
Assume that the horizontal homomorphism in case (2) is not injective. Let 0 satisfy the
equation
0 +
0
0 = 0. Then (1 +
0
(
1)
)
for some
(
1). Therefore(1 + 0)
(
1))
=
1 for some 1
(
1)+1 . Thus, (1 + 0) (
1)) 11
(
1) is a
primitive
th root of unity.
Let be a complete discrete valuation field.
Ifchar( ) = 0, then is an open subgroup in for 1. Ifchar( ) = 0, then
is an open subgroup in
if and only if
is relatively prime to
.
contains finitely many roots of unity of order a power of .
Proof. If char( ) = 0, then we get 1 for 1. It means that is open. If
char( ) = , then 1 for ( ) = 1 and is open. In this case, if char( ) =
,
then 1 +
for (
) = 1. Then
is not open. If char( ) = 0, then we obtain
when
( 1) + ( 1) . Therefore is open for 1.
This corollary demonstrates that for complete discrete valuation fields of characteristic 0 with
residue field of characteristic the topological properties are closely related with the algebraic
ones. The case char( ) = is very different.
-
7/26/2019 Lectures on Local Fields
7/57
7
(1.11). Product representation. Now we deduce a multiplicative analog of the expansion in
the corollary of (1.8).
Let be a complete discrete valuation field. Let be a set of
representatives. Then for there exist uniquely determined ,
for
0,
0 , such that can be expanded in the convergent product
=
0
1
(1 +
)
Proof. The existence and uniqueness of and
0 immediately follow. Assume that
,
then find
with (1 +
)
1
+1. Proceeding by induction, we obtain an
expansion of in a convergent product. If there are two such expansions (1 +
) =
(1 +
), then the residues
,
coincide in . Thus,
=
.
(1.12).
-Structure of The Group of Principal Units. Everywhere in this section is a
complete discrete valuation field with residue field of positive characteristic .
If 1 then
1 as + . This enables us to define
= lim
if lim
=
Let 1, . Then 1 is well defined and
+ =
,
= (
) ,
( )
=
for 1, . The multiplicative group 1 is a -module under
the operation of raising to a power. Moreover, the structure of the -module
1 is compatible
with the topologies of
and 1.
Proof. Assume that lim
= lim
; hence
0 as + and lim
= 1.
A map
1 1 ( ( ) ) is continuous with respect to the -adic topology on
and the discrete valuation topology on 1. This argument can be applied to verify the other
assertions of the lemma.
Let be of characteristic
with perfect residue field. Let be a set of
representatives, and let 0 be a subset of it such that the residues of its elements in form a
basis of as a vector space over
. Let an index-set numerate the elements of 0 . Let
be the -adic valuation.
Then every element 1 can be uniquely represented as a convergent product
=
( )=1 0
(1 +
)
0
and the sets = : ( ) are finite for all 0, (
) = 1.
Proof. We first show that the element can be written modulo
for 1 in the desired
form with . Proceeding by induction, it will suffice to consider an element
modulo
+1. Let 1 +
mod
+1 ,
. If ( ) = 1, then one can find
1
0 and 1
such that 1 +
=1(1 +
) mod
+1 for
some . If = with an integer , ( ) = 1, then one can find
1
0 and
-
7/26/2019 Lectures on Local Fields
8/57
8
1
such that 1 +
=1(1 +
)
mod
+1 for some . Now due
to the continuity we get the desired expression for 1 with the above conditions on the sets
.
Assume that there is a convergent product for 1 with
, . Let (
0 ) = 1 and 0 be
such that = (
0
0 )
0
(
) for all (
) = 1, . Then the choice of 0 imply
(1 +
)
+1, which concludes the proof.
The group 1 has a free topological basis 1 +
where where
0,
(
) = 1.
If = ( ) is divisible by 1, let
: be the map
+
0 . Then raising to
the th power in case (2) of the proposition in (1.10) is described by
.
Let be of characteristic 0 with perfect residue field of characteristic
.
Let be a set of representatives and let
0 (resp. 0 ) be a subset of it suchthat the residues
of its elements in form a basis of as a vector space over (resp. are -generators of
(
) ). Let the index-set
(resp.
)numerate the elements of
0 (resp.
0 ). Let =
:
1
( 1) (
) = 1
Then every element 1 can be represented as a convergent product
=
(1 +
)
(1 +
(
1))
0 0
(the second product occurs when ( 1) is an integer)and the sets
=
:
(
)
=
: (
)
are finite for all 0,
.
Proof. We shall show how to obtain the required form for
modulo
+1 . Let
= 1 +
mod
+1,
. There are four cases to consider:
(1) . One can find
1
0 and 1
satisfying the congruence
1 +
=1(1 +
) mod
+1 for some .
(2)
( 1), =
with . Then thereexist
1
0, 1
such that
1 +
=1
(1 +
)
mod
+1 for some
(3) (
1)
, =
(
1). The definition of 0 imply that if =
with
, then there exist
1
0, 1
0 , 1
, 1
such
that
1 +
=1
(1 +
)
=1
(1 +
)
mod
+1 for some
(4) ( 1). If = min : ( 1) and = , then
1 +
(1 +
)
mod
+1 for some
-
7/26/2019 Lectures on Local Fields
9/57
9
Now applyingthe arguments of the precedingcasesto 1+
, we can write 1 +
mod
+1
in the required form.
Let
be of characteristic 0 with perfect residue field of characteristic
.(1) If
does not contain nontrivial
th roots of unity then the representation in the proposition
is unique. Therefore the elements of the proposition form a topological basis of 1 .
(2) If contains a nontrivial th root of unity, let be the maximal integer such that
contains a primitive
th root of unity. Then the numbers
of the proposition
are determined uniquely modulo
. Therefore the elements of the proposition form a
topological basis of 1
1 .
(3) If the residue field of is finite then 1 is the direct sum of a free -module of rank
and the torsion part.
Proof. (1) If then all horizontal homomorphisms of the diagrams in the second
Proposition of (1.10) are injective.
(2) Argue by induction on . Write a primitive
th root
in the form
=
(1 +
)
(1 +
(
1))
and raise the right hand side to the
th power which demonstrates the non-uniqueness.
Now if
1 =
(1 +
)
(1 +
(
1))
then we deduce that =
=
with
-adic integers . Then
(1 +
)
(1 +
(
1))
is a th root of unity, and so is equal to
(1 +
)
(1 +
(
1))
1
for some integer . Now by the induction assumption all
1
1
are
divisible by
1. Thus, all are divisible by
.
(3) If the residue field of is finite then 1 is a module of finite type over the principal
ideal domain
, so by the structure theorem on such modules it is a direct sum of a free
module and a finite torsion module. If a primitive th root of unity is in , then the kernel of
is of order . Hence :
( ) = , since is finite. The cardinality of is equal to
= [
(
1)]
[[
(
1)]
].
-
7/26/2019 Lectures on Local Fields
10/57
10
2: Extensions of Complete Fields
(2.1). Henselian Property. Let be a commutative ring. For two polynomials
( ) =
+ 0, ( ) =
+ + 0 their resultant is the determinant of the matrix
1
0
1
0
. . . . . .
. . . . . .
1
0
1
0
1
0
. . . . . .
. . . . . .
1
0
This determinant (
) is zeroiff
and have a common root; in general (
) =
1 +
1
for some polynomials 1
1
[
]. If (
) =
=1(
),
(
) =
=1(
),then their resultant (
) is
( ). In particular, ( ( )) = ( ).
If
[
] then
(
)
. We shall use the following properties of the resultant:
if
mod [
] then (
)
(
1 ) mod
; if (
)
+1 then
[ ]
[ ] + [ ].
Let be a complete discrete valuation field with the ring of integers , the maximal ideal
, and the residue field . For a polynomial
( ) =
+
+
0 [ ] we will denote
the polynomial
+
+
0 by
( )
[
]. We will write
( )
( ) mod
if
( )
( )
[
].
Let
0 0
be polynomials over such thatdeg
= deg 0+ deg 0 and the
leading coefficient of
coincides with that of
0 0. Let
( 0 0)
+1
0 0 mod 2 +1
for an integer
0.
Then there exist polynomials (
)
(
) such that
=
deg = deg 0 deg = deg 0
0 mod +1
0 mod +1
Proof. We first construct polynomials ( ) ( ) [ ] with the following properties:
deg(
0) deg
0, deg( 0) deg 0
(
)
1( ) mod +
(
)
1( ) mod +
( ) ( ) ( ) mod +2 +1
Proceeding by induction, we can assume that the polynomials ( ) ( ), for
1,
have been constructed. For a prime element put
(
) =
1( ) + +
(
)
(
) =
1( ) + +
(
)
-
7/26/2019 Lectures on Local Fields
11/57
11
with
( )
( ) [ ], deg
( ) deg 0( ), deg
( ) deg 0( ). Then
(
)
(
)
1( )
1( ) +
1( )
(
) +
1( )
(
) mod
+2 +1
Since by the induction assumption
( )
1(
)
1(
) =
+2
1(
) for a suitable
1( ) [ ] of degree smaller than that of
, we deduce that it suffices for
(
),
(
)
to satisfy the congruence
1( )
1( )
( ) +
1( )
( ) mod
+1 .
However, (
1( )
1( )) (
0( ) 0( )) 0 mod +1. Then the properties
of the resultant imply the existence of polynomials
,
satisfying the congruence. Write
=
1 +
1 with polynomial
1 of degree smaller than that of
1 . Then it is easy
to see that the degree of
=
+
1 is smaller that the degree of
1 . The polynomials
are the required ones.
Now put ( ) = lim ( ) ( ) = lim ( ) and get
( ) = ( ) ( ).
The following statement is often called Hensel Lemma; it was proved by K. Hensel for -adic
numbers and byK. Rychlikfor complete valuation fields.
Let
0 0 be monic polynomials with coefficients in . Let
=
0 0 andsuppose that 0 0 are relatively prime in [ ]. Then there exist monic polynomials
with
coefficients in , such that
( ) = (
) (
)
( ) = 0( ) ( ) = 0( )
Proof. We have (
0( )
0( )) and we can apply the previous proposition for = 0.
The polynomials ( ) and ( ) may be assumed to be monic, as it follows from the proof of
the proposition.
Valuation fields satisfying the assertion of Corollary 1 are said to be Henselian. Corollary 1
demonstrates that complete discrete valuation fields are Henselian.
Let
( ) be a monic polynomial with coefficients in . Let
( 0)
2 +1
( 0)
+1
for some 0 and integer 0. Then there exists such that 0
+1 and
( ) = 0.
Proof. Put 0( ) = 0 and write
( ) =
1( )( 0) + with . Then
2 +1. Put 0( ) =
1( ) [ ]. Hence
( ) 0( ) 0( ) mod 2 +1 and
(
0) = 0( 0) +1. Hence ( 0( ) 0( ))
+1, and the proposition implies the
existence of polynomials ( ) ( ) [ ] such that ( ) = 0 mod +1 ,
and
( ) = ( ) ( ).
If the residue field of is finite , then for every
the polynomial
(
) =
1
1 has a root
such that
=
. So one hasmultiplicative representativesof
consisting of0 and all roots of unity in of order dividing
1 (or, equivalently, all roots
of unity in of order prime to ).
For every positive integer there is such that 1 + .
-
7/26/2019 Lectures on Local Fields
12/57
12
Proof. Put
( ) = with 1+ . Let
+1. Then
(1)
+1 .
Therefore for every 1 + 2
+1the polynomial
( ) has a root 1 mod
+1 due to
Corollary 2.
Let
( ) = +
1 1 + + 0 be an irreducible polynomial with coefficients
in . Then the condition
0 implies for 0
1.
Proof. Assume that 0 and that is the maximal such that ( ) = min0
1 ( ).
If , then put
1( ) = 1
( )
0( ) =
+ 1
1
1 + + 1
0
0( ) = 1
+ 1
We have
1( ) =
0( ) 0( ), and
0( ) 0( ) are relatively prime. Therefore, by the
proposition
1( ) and
( ) are not irreducible.
(2.2). and
. Let be a field and an extension of with a valuation : .
Then induces the valuation 0 = : on . In this context is said to be
an extension of valuation fields. The group 0( ) is a totally ordered subgroup of ( )
and the index of 0( ) in ( ) is called the ramification index ( ). The ring of
integers 0
is a subring of the ring of integers and the maximal ideal
0coincides with
0. Hence, the residue field
0can be considered as a subfield of the residue field
. Therefore, if is an element of
0, then its residue in the field
0can be identified
with the image of as an element of in the field . We shall denote this image of
by . The degree of the extension 0
is called the intertia degree
(
). An
immediate consequence is the following lemma.
Let
be an extension of
and let
be a valuation on
. Let
andlet 0 be the induced valuation on . Then
( ) = ( ) ( 0)
(
) =
(
)
( 0)
(2.3). Extension of Discrete Valuation. Assume that is a finite extension and 0 is
a discrete valuation. Let elements 1
for natural ( ) be such that
( 1) + ( )
(
) + (
) are distinct in ( ) ( ). If =1 = 0 holds
with , then, as ( ) are all distinct, we get =1
= min1
(
) and
so = 0 for 1
. This shows that 1
are linearly independent over and
hence ( ) is finite. Let be a prime element with respect to 0 . Then we deduce that
there are only a finite number of positive elements in ( ) which are ( ). Consider the
smallest positive element in ( ). It generates the group ( ), and we conclude that is
a discrete valuation. Thus, we have proved the following result.
Let be a finite extension and 0 discrete for a valuation on . Then is
discrete.
-
7/26/2019 Lectures on Local Fields
13/57
13
(2.4).
for Complete Fields. Let and be fields with discrete valuations and
respectively and . The valuation is said to be an extension of the valuation , if
0 = for a positive . We shall write and use the notations ( )
( ) instead of
(
)
(
). If then ( ) = ( ) ( ).
Let be a finite extension of of degree ; then
(
)
( )
Proof. Let = ( ) and let
be a positive integer such that
( ). Let
1
be elements of such that their residues in are linearly independent over
. It suffices
to show that
are linearly independent over for 1
0 1. Assume
that
= 0
for and not all = 0.
Multiplying the coefficients by a suitable power of , we may assume that
and not all . Note that if
, then
= 0 and so .Therefore, there exists an index such that
. Let 0 be the minimal such index.
Then 0 = (
), which is impossible. We conclude that all = 0. Hence,
and ( )
( )
.
Let be the completion of with the discrete valuation . Then ( ) =
1
( ) = 1. Note that if is not complete, then : = ( )
( ). On the contrary, in
the case of complete discrete valuation fields we have
Let be an extension of and let be complete with respect to discrete
valuations
. Let
=
( ) and
=
(
)
. Let
be a prime element
with respect to and
1
elements of such that their residues form a basis of
over
. Then
is a basis of the
-space
and of the
-module
, with1
0 1. If
, then is a finite extension of degree =
.
Proof. Let be a set of representatives for . Then the set
=
=1
:
and almost all = 0
is the set of representatives for . For a prime element with respect to put
=
,
where = + 0 . Using Corollary (1.8) we obtain that an element can be
expressed as a convergent series
=
with
Writing
=
=1
with
-
7/26/2019 Lectures on Local Fields
14/57
14
we get
=
+
Thus, can be expressed as
with
=
+
1
0
1
By the proof of the previous lemma this expression for is unique. We conclude that
form a basis of over and of over .
(2.5). Uniqueness of Extension of Discrete Valuation From a Complete Field. Further we
shall assume that ( ) = for a discrete valuation . Then ( ) = : ( ) for an
extension of .
Let be a complete field with respect to a discrete valuation and a finite
extension of . Then there is precisely one extension on of the valuation
and =
1
with
=
( ). The field is complete with respect to .
Proof. Let =
. First we verify that is a valuation on . It is clear that
(
) = +
if and only if = 0 and ( ) = ( ) + ( ). Assume that ( ) ( )
for , then
( +
) = (
) +
1 +
and it suffices to show that if ( ) 0, then (1 + ) 0. Let
( ) =
+
1 1 +
+
0
be the monic irreducible polynomial of over . Then we get ( 1) 0 = ( )
( ) and if
=
: ( )
, then (( 1) 0
)
=
( ). We deduce that ( 0
) 0, and making use of
(2.1), we get ( ) 0 for 0
1. Now
( 1)
( )
(1 + ) =
( 1) = (
1) +
1( 1) 1 +
+
0
hence
( )
(1 + ) 0 and
(1 + ) 0
i.e., (1 + ) 0. Thus, we have shown that is a valuation on .
Let = : ; then ( ) = ( ) for . Hence, the valuation (1 ) is an
extension of to (note that (1 ) ( ) = in general). Let = ( (1 ) ); is
finite. Put = ( ) : , hence ( ) = ( ) = with a prime element
with respect to . Therefore, = ( )
is at once a discrete valuation on and an
extension of .
Let 1
be a basis of the -vector space . By induction on , 1 , we
shall show that
=1
(
)
0
(
)
0
for
= 1
-
7/26/2019 Lectures on Local Fields
15/57
15
where (
)
.
The left arrow and the case = 1 are clear. For the induction step we can assume that
(
)
0 for each
= 1
. Therefore we can assume that ( (
)
) is bounded for
= 1 . Hence
1+
=2
(
)
=
(
)1
1
=1
(
)
0
where (
)
= (
(
)1
1
(
)
. Then
=2
( (
)
(
+1)
) =
=2
(
)
=2
(
+1)
0
and the induction hypothesis shows that (
)
(
+1)
0 for
= 2
. Thus, each
(
(
)
)
converges to, say,
. Finally, the sequence 1+
=2 (
)
converges both to
0 and to 1+
=2 , so 0 = 1+
=2 which contradicts the choice of .
Similarly one shows that a sequence
=1
(
)
is fundamental if and only if
(
)
is
fundamental for each
= 1
.
Thus, the completeness of implies the completeness of its finite extension with respect
to any extension of . We also have the uniqueness of the extension.
Let be a finite Galoisextension of . Then
= for the discrete valuation
on and
Gal( ). If is a prime element in , then
is a prime element and
=
,
=
.
Let be a finite extension. Let
and let
( ) be the monic irreducible
polynomial of over
. Then
( )
[
]. Conversely, let
( ) be a monic polynomial
with coefficients in . If is a root of
( ), then .
Proof. It is well known that =
is separable over for some 0. Let be a
finite Galois extension of with . Then, in fact, and the monic irreducible
polynomial ( ) of over can be written as
( ) =
=1
(
)
Gal( ) 1 = 1
By the previous corollary we get
. Hence we obtain ( )
[ ] and
( ) =
[ ]. If is a root of the polynomial
( ) = +
1 1 + + 0
[
] and , then1 =
1 1
0 , contradiction. Thus, .
(2.6). Types of Ramified Extensions. From now on is a complete discrete valuation field.Let be an algebraic extension. If
is the unique discrete valuation on which
extends the valuation = on , then we shall write ( )
( ) instead of (
),
(
). We shall write or or
or or for the ring of integers
-
7/26/2019 Lectures on Local Fields
16/57
16
, the maximal ideal
, the group of units
, a prime element with respect to , and
the residue field , respectively.
A finite extension of a complete discrete valuation field is called unramifiedif
is a separable extension of the same degree as . A finite extension is called totally
ramifiedif
( ) = 1. A finite extension is called tamely ramifiedif is a separable
extension and
( ) where = char( ) 0. A finite extension is calledtotally
ramifiedif = . A finite extension is calledwildly totally ramifiedif = and the
degree of is a power of =char( ).
(2.7). Unramified Extensions. If is unramified then we deduce From Lemma (2.4) that
(
) = 1,
( ) =
:
.
(1) Let be an unramified extension, and = (
) for some
. Let
be such
that =
. Then = (
), and is separable over
,
=
[ ];
is a simple
root of the polynomial
( ) irreducible over
, where
( ) is the monic irreducible
polynomial of over
.
(2) Let
( ) be a monic polynomial over , such that its residue is a monic separablepolynomial over
. Let
be a root of
( ) in
alg, and let = (
). Then the
extension is unramified and =
(
) for
= .
Proof. (1) By the preceding lemma
( )
[
]. We have
( ) = 0 and
( ) = 0,
deg
( ) = deg
( ). Furthermore,
:
(
) :
= deg
( ) = deg
( )
(
) :
=
:
It follows that = ( ) and
is a simple root of the irreducible polynomial
( ). Therefore,
(
) = 0 and
( ) = 0, i.e., is separable over . Thus,
= [ ].
(2) Since the residue of
is separable, it is separable too. Let
( ) =
=1
(
) be the
decomposition of
( ) into irreducible monic factors in [ ]. Then every root of
( ) is in
the ring of integers of any finite Galois extension of which contains all of them; and therefore
(
)
[
]. Suppose that is a rootof
1( ). Then
1( ) =
1( ) is a monic separable
polynomial over . The Henselian property of implies that
1( ) is irreducible over .
We get
. Since
=
, we obtain (
) and
deg
1( ) = : : (
) :
= deg 1( ) = deg
1( )
Thus, = (
), and is unramified.
(1) If are unramified, then is unramified.
(2) If is unramified, is an algebraic extension of and is the discrete valuation
field with respect to the extension of the valuation of , then
is unramified.
(3) If
1
2 are unramified, then
1
2 is unramified.
Proof. (1) follows from the multiplicativity of the ramification index. To verify (2) let
= (
) with
,
( )
[
] as in the first part of the proposition. Then
because = ( ). Observing that = ( ), we denote the irreducible monic polynomial
-
7/26/2019 Lectures on Local Fields
17/57
17
of over by
1( ). By the Henselian property of we obtain that
1( ) is a power of
an irreducible polynomial over . However,
1( ) divides
( ), hence
1( ) is irreducible
separable over . Applying the second part of the proposition, we conclude that is
unramified.
(3) follows from (1) and (2).
An algebraic extension of is calledunramifiedif are separable extensions
and ( ) = 1, where is the discrete valuation on , and is the unique extension of
on .
The third assertion of the Corollary shows that the compositum of all finite unramified
extensions of in a fixed algebraic closure alg of is unramified. This extension is not
a complete field in general, but a Henselian discrete valuation field. It is called the maximal
unramified extension ur of . Its maximality implies
ur = ur for any automorphism of
the separable closure sep over . Thus, ur is Galois.
(1) Let be an unramified extension and let be a Galois extension. Then is
Galois.
(2) Let be an unramified Galois extension. Then is Galois. For an automorphism
Gal(
) let
be the automorphism in Gal( ) satisfying the relation
=
for every
. Then the map
induces an isomorphism of Gal( ) onto
Gal( ).
Proof. (1) It suffices to verify the first assertion for a finite unramified extension . Let
= (
) and let ( ) be the irreducible monic polynomial of
over . Then
( ) =
=1
(
)
with
1 =
. Let
( ) be a monic polynomial over of the same degree as ( )
and
(
) =
(
). The Henselian property implies
( ) =
=1
(
)
with
=
. The first proposition above shows that = ( 1), and we deduce that
is Galois.
(2) Note that the automorphism
is well defined. Indeed, if
with = , then
(
)
and
=
. It suffices to verify the second assertion for a finite unramified
Galois extension . Let
( ) be as in the first part of the first proposition. Since all
roots of
( )belong to , we obtain that allroots of
( ) belong to and is Galois. The
homomorphism Gal( ) Gal( ) defined by
is surjective because the condition
=
implies
=
for the root of
( ) with =
. Since Gal( ) Gal( )
are of the same order, we conclude that Gal(
) is isomorphic to Gal(
).
The residue field of ur coincides with the separable closure sep
of and
Gal(
ur
) Gal(
sep
).
-
7/26/2019 Lectures on Local Fields
18/57
18
Proof. Let
sep, let ( ) be the monic irreducible polynomial of
over , and
( ) as
in the second part of the first proposition. Let be all the roots of
( ) and = ( ).
Then ur and
=
ur for a suitable
. Hence, ur = sep
.
(2.8). Maximal Unramified Field. Let be an algebraic extension of , and let be a
discrete valuation field with perfect residue field. We will assume that alg = alg.
Let be an algebraic extension of and let be a discrete valuation field.
Then ur =
ur, 0 = ur is the maximal unramified subextension of
which is
contained in , and 0 is totally ramified.
Proof. We have ur ur. Since the residue field of ur coincides with the residue field
of ur, we deduce ur = ur. An unramified subextension of in is contained in 0 ,
and 0 is unramified.
Let be a finite Galoisextension of and let 0 be the maximal unramified
subextension in
. Then
0
and
0
are Galois, and the map
defined in thesecond proposition of(2.7)induces the surjective homomorphism
Gal( )
Gal( 0 ) Gal( 0 ) = Gal( )
The extension ur
is Galois and Gal( ur 0) Gal(
ur
) Gal( ur
ur),
Gal( ur
ur) Gal( 0), Gal(
ur
) Gal(
ur 0).
Proof. Let
Gal(
). Then
0 is unramified over , hence 0 = 0 and 0
is Galois. The surjectivity of the homomorphism Gal( ) Gal( 0 ) follows from the
second proposition of (2.7). Since and
ur
are Galois extensions, we obtain that
ur is a Galois extension. Then ur = ur by the previous proposition. The remaining
assertions are easily deduced by Galois theory.
(2.9). Tamely Ramified Extensions.
(1) Let be a finite tamely ramified extension of , and let 0 be the maximal unramified
subextension in . Then = 0( ) and = 0 [ ] with a prime element in
satisfying the equation 0 = 0 for some prime element 0 in 0, where = ( ).
(2) Let 0 be a finite unramified extension, = 0( ) with = 0. Let
if
= char( )
0. Then
is separable tamely ramified.
Proof. (1) (2.8) shows that 0 is totally ramified. Let 1 be a prime element in 0 , then
1 =
for a prime element
in and
. Since = 0 , there exists 0 such
that = . Hence 1 1 =
for the principal unit = 1
. For the polynomial
( ) = we have
(1)
,
(1) = . Now Corollary 2 of (2.1) shows the existence
of an element
with
=
,
= 1. Therefore,
= 1
1
, 0 =
are the elementsdesired for the first part of the Proposition.
(2) Let = 1 for a prime element 1 in 0 and a unit 0 . The polynomial
( ) =
is separablein 0[ ] and we canapply the Henselianproperty to
( ) =
-
7/26/2019 Lectures on Local Fields
19/57
19
and a root
sep of
( ). We deduce that 0( ) 0 is unramified and hence it suffices
to verify that 0 for = ( ) 0 = 0( ), is tamely ramified. We get = 0( 1)
with 1 = 1 , 1 = 1 . Put = g.c.d.( ). Then 0( 2 ) with
2 =
1
and a primitive th root of unity. Since the extension 0
( )
0 is unramified (this can
be verified by the same arguments as above), 1 is a prime element in 0( ). Let be the
discrete valuation on 0( 2 ). Then ( ) ( 1) ( ) and ( 1) ( ) , because
and are relatively prime. This shows that 0( 2 ) 0( )
. However,
0( 2) : 0( ) , and we conclude that 0( 2) 0( ) is tamely and totally
ramified. Thus, 0( 2) 0 and 0 are tamely ramified extensions.
(1) If are tamely ramified, then is separable tamely ramified.
(2) If is tamely ramified, is an algebraic extension, and is discrete, then
is tamely ramified.
(3) If 1 2 are tamely ramified, then 1 2 is tamely ramified.
Proof. It is carried out similarly to (2.8). To verify (2) one can find the maximal unramified
subextension 0 in . Then it remains to show that 0 is tamely ramified. Put
= 0( ) with = 0. Then we get = 0( ), and the second part of the proposition
yields the required assertion.
(2.10). Totally Ramified Extensions. A polynomial
( ) = +
1 1 + + 0 over
is called an Eisenstein polynomialif
0
1 0 2
(1) The Eisenstein polynomial
( ) is irreducible over . If is a root of
( ), then
(
)
is a totally ramified extension of degree
,
is a prime element in
(
),
( )
=
[ ].
(2) Let be a separable totally ramified extension of degree , and let be a prime
element in . Then is a root of an Eisenstein polynomial over of degree .
Proof. (1) Let be a root of
( ), = ( ), = ( ). Then
( ) =
1
=0
min
0
1(
(
) +
( ))
where and
are the discrete valuations on and . It follows that
( )
0. Since
(
0) ( ) +
( ) for
0, one has
( ) =
(
0) = .Then ( ) = 1 =
= 1, and
=
[ ].
(2) Let be a prime element in . Then =
(
). Let
( ) =
+
1 1 +
+
0
be the irreducible polynomial of over . Then = and
( ) = min0
1 ( ) +
, hence ( ) 0, and = ( 0), ( 0) = 1.
-
7/26/2019 Lectures on Local Fields
20/57
20
(2.11). Ramification Groups. Let be a finite Galois extension of ,
= Gal( ). Put
=
:
+1
for all
1
Then
1 =
by Lemma (2.11) and
+1 is a subset of
.
Let
be the discrete valuation of . For a real number define
=
:
(
) + 1 for all
Certainly each of
is equal to
with the least integer
.
are normal subgroups of
.
Proof. Let
. Then
+1
. Hence
1( )
1(
+1
) =
+1
by Lemma (2.11), i.e.,
1
. Let
. Then
( )
=
( ( )
) +
( )
+1
i.e.,
. Furthermore, let
. Then ( )
for
and
( )
+1
, 1
( )
+1
, 1
.
The groups
are called (lower)ramification groupsof
= Gal( ).
Let be a finite Galois extension of , and let be a separable extension of
. Then
0 = Gal( 0) and the
th ramification groups of
0 and
coincide for
0.
Moreover,
=
0 : +1
for a prime element in , and
=
1
for sufficiently large
.
Proof. Note that
0 if and only if Gal( ) is trivial. Then
0 coincides with
the kernel of the homomorphism Gal( ) Gal( ). The first proposition of (2.11) and
the second proposition of (2.7) imply that this kernel is equal to Gal( 0). Since
is a
subgroup of
0 for
0, we deduce the assertion about the
th ramification group of
0 . We
get = 0 [ ]. Let =
=0
be an expansion of with coefficients in 0 .As
= for
0 it follows that
=
=0
(
)
Now we deducethe description of
, since
(
)
. If
max
(
) :
,
then
=
1
.
The group
0 is called theinertia group of
, and the field 0 is called the inertia subfield
of .
Let be a finite Galois extension of , a separable extension of
, and
a prime element in . Introduce the maps
0:
0
:
(
0)
by the formulas
(
) =
(
). Then
is a homomorphism with the kernel
+1 for
0.
-
7/26/2019 Lectures on Local Fields
21/57
21
Proof. The proof follows from the congruence
( )
=
mod
+1
for
. The kernel of
consists of those automorphisms
, for which
1 +
+1
, i.e.,
+2
.
Let be a finite Galois extension of , and a separable extension of .
If char( ) = 0, then
1 = 1 and
0 is cyclic. If char( ) =
0, then the group
0
1
is cyclic of order relatively prime to
,
+1 are abelian
-groups, and
1 is the maximal -subgroup of
0.
Proof. The previous proposition permits us to transform the assertions of this corollary into
the following: a finite subgroup in is cyclic (of order relatively prime to char( ) when
char( )
= 0 ); there are no nontrivial finite subgroups in the additive group of ifchar( ) = 0;
ifchar( ) = 0 then a finite subgroup in is a -group.
Let be a finite Galois extension of and a separable extension of
.
Then the group
1 coincides with Gal( 1), where 1 is the maximal tamely ramified
subextension in .
Proof. The extension 1 0 is totally ramified by the first proposition of (2.11) and is the
maximal subextension in 0 of degree relatively prime with char( ). Now Corollary 1
implies
1 = Gal( 1).
Let be a finite Galois extension of and a separable extension of
.
Then
0 is a solvable group. If, in addition, is a solvable extension, then is solvable.
Proof. It follows from Corollary 1.
(2.12). The Norm Map for Cyclic Extensions. Let be a local field and its Galois
extension of prime degree . Then there are four possible cases:
is unramified;
is tamely and totally ramified;
is totally ramified of degree = char( ) 0;
Let be a separable extension of prime degree ,
. Then
(1 + ) = 1 +
( ) + Tr
( ) + Tr
( )
with some
such that
( ) 2
( ) (
and Tr
are the norm and the trace
maps, respectively).
Proof. Recall that for distinct embeddings
of over into the algebraic closure of ,
1
, one has
=
=1
( )
Tr
=
=1
( )
-
7/26/2019 Lectures on Local Fields
22/57
22
Hence
(1 + ) =
=1
(1 +
( )) = 1 +
=1
( ) +
=1
1
( ) + +
=1
( )
For = 1
( ) +
we get
( ) 2
( ).
Our nearest purpose is to describe the action of the norm map
with respect to the
filtration on and .
Let be an unramified extension of degree . Then a prime element
in is a prime element in . Let
= 1 +
,
= 1 +
. Then the following
diagrams are commutative:
0
0
Tr
Proof. The commutativity of the first two diagrams is easy. The preceding Lemma shows that
(1 +
) = 1 + (Tr
)
+ (
)
+ Tr
( )
with
( ) 2
and, consequently, Tr
( ) 2
. Thus, we get
(1 +
) 1 + (Tr
)
mod +1
and the commutativity of the third diagram.
In the case under consideration
1 = 1 .
Let be a totally and tamely ramified cyclic extension of degree . Thenfor some prime element
in , the element =
is prime in and = . Let
= 1 +
,
= 1 +
. Then the following diagrams
id
0
0
=
are commutative, where id is the identity map,
takes an element to its th power, is
the multiplication by ,
1. Moreover,
=
+1 if
.
Proof. Since
=
and is Galois, then Gal( ) is cyclic of order and
(
) =
for a generator
ofGal( ), where is a primitive th root of unity, .It is easy to see that the first diagram is commutative.
Corollary in (2.5) shows that
( ) =
for
Gal(
),
, and we get the
commutativity of the second diagram.
-
7/26/2019 Lectures on Local Fields
23/57
23
If =
, then 1 +
for
, and
(1 +
) = (1 +
) 1 +
mod
+1
We deduce
=
=
Finally, 1 = 1
=0 (
), therefore for
and for
one has
(1 +
) =
1
=0
(1 +
) = 1 (
)
Thus
=
+1 .
In the case under consideration
1 = 1 . If is algebraically closed
then
= .
Now we treat the most complicated case where
is a totally ramified Galois extensionof degree = char( ) 0. In this case
=
[
], = (
) for a prime element
in , and = . Let
be a generator of Gal( ), then
(
)
. One can write
(
)
=
with
1 +
. Then
2(
)
=
(
)
=
2
( )
and
1 =
(
)
=
( )
1( )
This shows that
1 +
and
1 +
, because raising to the th power is an injective
homomorphism of . Thus, we obtain
(
)
1 +
. Put
(
)
= 1 +
with
= (
)
1 ( )
Note that does not depend on the choice of the prime element
and of the generator
of
= Gal( ). Indeed, we have
(
)
1 +
mod
+1
and
( )
1 mod
+1
for an element
. We also deduce that
( )
for every element . This means that
=
,
+1 = 1 .
Let
( ) =
+
1
1 + +
0 be the irreducible polynomial of over .
Then
Tr
(
) =
0 if 0 2
1 if
= 1
-
7/26/2019 Lectures on Local Fields
24/57
24
Proof. Since
(
) for 0
1 are all the roots of the polynomial
( ), we get
1
( )
=
1
=0
1
(
)
(
)
Putting = 1 and performing the calculations in the field (( )), we consequently deduce
( ) =
(1 +
1 + + 0
)
1
( )
=
1 +
1 + + 0
mod +1
1
(
) =
1
(
) =
0
(
) +1
(because 1 (1 ) =
0
in (( )) ). Hence
0
1
=0
(
) +1
( )
mod +1
or
Tr
(
) =
1
=0
(
)
(
) =
0 if 0
2
1 if = 1
as desired.
Let [ ] denote the maximal integer
. For an integer
0 put
(
) =
+ 1 + (
1
)
. Then
Tr
(
) =
( )
Proof. One has
(
) =
1 =1
(
) and
(
)
1 +
mod
+1
.
Then
(
) = ( 1)!( )
1
(
1)( +1)
with some 1 + (
1)( +1)+1
. Since = , for a prime element in one has the
representation =
with
. The previous lemma implies
Tr
+ +1
+ +1
=
0 if 0 1
+1
if = 1
for + +1 = ( )
+1
( 1)!(
)
1
. Since Tr
(
) =
Tr
( ) we can choose
the units
+
+1 , for every integer
, such that Tr
(
+ +1
+
+1) = 0 if
(
+ 1) and=
+( +1)
if ( + 1). Thus, since the -module
is generated by
,
, we
conclude that Tr
(
) = ( )
.
-
7/26/2019 Lectures on Local Fields
25/57
25
Let be a totally ramified Galois extension of degree = char( ) 0.
Let
be a prime element in . Then =
is a prime element in . Let
=
1 +
= 1 +
. Then the following diagrams are commutative:
id
0
0
=
=
1
+
+
=
(
1)
+
+
where 1
in the third diagram and
0 is the last diagram.
Moreover,
(
+
) =
(
+ +1 ) for
0
.
Proof. The commutativity of the first and the second diagrams is easy.
To treat the remaining diagrams, put = 1 +
with
. Then, by the first lemma,
we get
= 1 +
(
)
+ Tr
(
) + Tr
(
)
with
( ) 2
. The previous proposition implies that
Tr
(
) + 1 +
1
Tr
( ) + 1 +
2
1
and for
Tr
(
)
+ 1 Tr
( )
+ 1
Therefore, the third diagram is commutative.
Further, using ( ), write
1 =
(
)
1 +
( )
+ Tr
(
) mod
+1
We deduce that
Tr
(
)
( )
mod
+1
Since
( )
mod
in view of
1 , we conclude that
(1 +
) 1
(
)
+1
for
. This implies the commutativity of the fourth (putting
) and the fifth (when
) diagrams.Finally, if
, then
(1 +
)
1 +
1 +
+
mod
+ +1
-
7/26/2019 Lectures on Local Fields
26/57
26
This means that
(1 +
+
)
+ +1 and
( + ) =
( + +1 ).
Compare the behaviour of the norm map with the behaviour of raising to the
thpower in Proposition (1.10).
+1 =
+1 .
If is algebraically closed then
= .
Proof. It follows immediately from the last diagram of the proposition, since the multiplication
by ( )
1 is an isomorphism of the additive group .
3: Local Class Field Theory
This section focuses on complete discrete valuation fields with finite residue field.
3.1. Useful Results on Local Fields with Finite Residue Field
(3.1.1). Structures. Let be a local field with finite residue field = ,
=
elements.
Since char( ) = , is of characteristic 0 or of characteristic .
In the first case ( ) 0 for the discrete valuation in , hence the restriction of on
is equivalent to the -adic valuation (by Ostrowskis Theorem). Then we can view the field
of -adic numbers as a subfield of . Let = ( ) = ( ) be the absolute ramification index
of . Then is a finite extension of of degree =
. Such a field was called a local
number field.
In the second case is isomorphic (with respect to the field structure and the discrete
valuation topology) to the field of formal power series (( )) with prime element , since
the multiplicative representatives of the residue field form a finite subfield of . Such a field
was called a local functional field.
is a locally compact topological space with respect to the discrete valuation
topology. The ring of integers and the maximal ideal
are compact. The multiplicative
group is locally compact, and the group of units is compact.
Proof. Assume that is not compact. Let ( ) be a covering by open subsets in , i.e.,
= , such that isnt covered by a finite union of . Let be a prime element of .
Since is finite, there exists an element
0 such that the set
0+ is not contained
in the union of a finite number of . Similarly, there exist elements
1
such that
0+
1 + +
+ +1 is not contained in the union of a finite number of . However,the element
= lim
+
=0
belongs to some
, a contradiction. Hence,
is
compact and , as the union of
+ with
= 0, is compact.
-
7/26/2019 Lectures on Local Fields
27/57
27
(3.1.2). Galois Extensions.
The Galois group of every finite extension of is solvable.
Proof. Follows from (2.11).
For every
1 there exists a unique unramified extension of of degree
: = (
1). The extension is cyclic and the maximal unramified extension ur of
is a Galois extension. Gal(
ur
) is isomorphic to and topologically generated by an
automorphism such that
( )
mod
ur for ur
The automorphism is called the Frobenius automorphism of
.
Proof. First we note that, by (2.1) contains the group
1 of ( 1) th roots of unity
which coincides with the set of nonzero multiplicative representatives of in . Moreover,
the unit group is isomorphic to
1 1 .
The field
has the unique extension
of degree
, whichis cyclic over
. (2.7) showsthat there is a unique unramified extension of degree over and hence = (
1).
Now let be an unramified extension of and . Then ( ) is of finite degree.
Therefore, ur is contained in the union of all finite unramified extensions of . We have
Gal(
ur
) lim
Gal(
)
It is well known that Gal( sep
) is topologically generated by the automorphism
such
that
( ) =
for sep
. Hence, Gal( ur ) is topologically generated by the Frobenius
automorphism .
If
1 , then
(
)
mod
and (
)
1. The uniqueness of the multiplicative representative for
implies
now that (
) =
.
(3.1.3).
Let
be a primitive th root of unity. Put
(
)
=
(
). Then
(
)
(
) = 0
and
belongs to the ring of integers of ( )
. Let
( ) =
1
1 1
=
(
1) 1
+
(
2) 1
+ + 1
Then
is a root of
( ), and hence ( )
:
( 1) 1. The elements
,
0
, are roots of
( ). Hence
( ) =
0
(
) and =
(1) =
0
(1
)
-
7/26/2019 Lectures on Local Fields
28/57
28
However,
(1
)(1
) 1 = 1 +
+ +
1
belongs to the ring of integers of ( )
. For the same reason, (1
)(1
)
1 belongs to
the ring of integers of ( )
. Thus, (1
)(1
)
1 is a unit and = (1
)
1(
1)
for some unit . Therefore, ( ( )
)
(
1)
1, and ( )
is a cyclic totally ramified
extension with the prime element 1
, and of degree ( 1) 1 over . In particular,
(
)
=
[1
] =
[
]
(3.1.4). The Group of Principal Units. If char( ) = , then Proposition (6.2) Ch. I shows
that every element 1 can be uniquely expressed as the convergent product
=
0
(1 +
)
where the index-set numerates
elements in , such that their residues form a basis of
over , and the elements
belong to this set of
elements; are elements of with
(
) =
, and . Thus, 1 has the infinite topological basis 1 +
.
Now let char( ) = 0. Every element 1 can be expressed as a convergent product
=
(1 +
)
where = 1
(
1),
, = ( ); the index-set numerates
elements in
, such that their residues form a basis of over , and the elements
belong to this set
of
elements; are elements of with ( ) =
, and .
If a primitive th root of unity does not belong to , then = 1 = 0 and the above
expression for is unique; 1 is a free -module of rank =
= :
.
If a primitive th root of unity belongs to , then = 1 +
(
1) is a principal unit
such that
, and
. In this case the above expression for
is not unique.
1 isisomorphic to the product of copies of and the -torsion group
, where 1 is the
maximal integer such that
.
If char( ) = 0, then is an open subgroup of finite index in for 1. If
char( ) = , then is an open subgroup of finite index in for . If char( ) =
and
, then is not open and is not of finite index in .
Proof. It follows (1.9) and (1.10) and the previous considerations.
(3.1.5). The Norm Groups. Now we have a look at the norm group
( ) for a finite
extension of . Recall that the norm map
:
is surjective when
.Let be a finite unramified extension. Then (2.12) implies that
= in the
case of an unramified extension and
= , where is a prime element
in , = : . This means, in particular, that
is a cyclic group of order
-
7/26/2019 Lectures on Local Fields
29/57
29
in this case. Conversely, every subgroup of finite index in
that contains
coincides with
the norm group
for a suitable unramified extension .
Let be a totally and tamely ramified Galois of degree . (2.12) shows that
1
=
1
for a suitable prime element in (e.g. such that = (
), and
for
if and only if
. Since is Galois, we get
and (
1).
Hence, the subgroup
is of index in
, and the quotient group
is cyclic. We
conclude that
=
1
with an element
, such that its residue
generates
. So
is
cyclic of order . Conversely, every subgroup of index relatively prime to char( ) coincides
with the norm group
for a suitable cyclic extension .
Let be a totally ramified Galois extension of degree
. The right vertical
homomorphism of the fourth diagram in the last proposition of (2.12)
1
has
kernel of order
; therefore its cokernel is also of order
. Let
be such that
does not belong to the image of this homomorphism. Since is perfect, we deduce that
1 +
1 . The other diagrams imply that
is a cyclic group of order
and generated by 1 +
mod
. If char( ) = 0, then
(
1), where
=
(
), and if then = ( 1) and a primitive th root of unity belongs to ,
and = (
) for a suitable prime element in . In this case
is generated
by mod
.
(3.1.6). Completion of ur . The field ur is a Henselian discrete valuation field with
algebraically closed residue field and its completion is a local field with algebraically closed
residue field sep
.
Let be the set of multiplicative representatives of the residue field of if its characteristic
is . is the union of all sets
1 1 (which coincides with the set of all roots of unity
of order relatively prime to
) and of 0.Let be a finite separable extension of . Since the residue field of is algebraically
closed, is totally ramified.
The norm maps
:
:
are surjective.
Proof. Since the Galois group of is solvable, it suffices to consider the case of a Galois
extension of prime degree . Certainly, the norm of a prime element of is a prime element of
. The surjectivity of the norm maps follows from (2.12).
(3.1.7). Augmentation Group (
) .
For a finite Galois extension denote by ( ) the subgroup of 1
generated by 1 where runs through all elements of 1 and runs through all elements
ofGal( ).
-
7/26/2019 Lectures on Local Fields
30/57
30
Every unit in
can be factorized as
with
, 1 Since
1 = 1
we deduce that ( ) coincides with the subgroup of
generated by 1 ,
,
Gal(
).
Let be a finite Galois extension of . For a prime element of define
: Gal( )
(
)
(
) =
1 mod (
)
The map
is a homomorphism which does not depend on the choice of . It induces a
monomorphism
: Gal( )ab
(
) where for a group
the notation ab stands
for the maximal abelian quotient of
.
The sequence
1 Gal( )ab
( )
1
is exact.
Proof. Since 1 belongs to
, we deduce that ( 1) 1 ( ) and
1
1
1
mod
(
)
Thus, the map
is a homomorphism. It does not depend on the choice of , since ( ) 1
1 mod ( ).
Surjectivity of the norm map has already been proved.
Suppose first that Gal( ) is cyclic with generator
. The kernel of
coincides
with 1 . Since
is a homomorphism, we have
1
(
1)
mod (
). So
we deduce that 1 is equal to the product of ( ) and the image of
. This shows the
exactness in the middle term.
Note that
1 = ( 1+ + +
1) 1, so ( ) = 1
. If
1
( ), then
(
1)
=
1 for some
. Hence 1 belongs to and therefore : divides
and
= 1. This shows the injectivity of
.
Now in the general case we use the solvability of Gal( ) and argue by induction. Let
be a Galois cyclic subextension of such that = = . Put =
.Since
:
is surjective, we deduce that
(
) =
(
).
Let
= 1 for
. Then by the induction hypothesis there is Gal( )
such that
=
1
with ( ). Write =
with
(
). Then
1
1
belongs to the kernel of
and therefore by the induction hypothesis can be
written as 1 with
Gal( ), ( ). Altogether, 1 mod ( )
which shows the exactness in the middle term.
To show the injectivity of
assume that 1 ( ). Then 1
(
) and
by the previous considerations of the cyclic case
acts trivially on . So
belongs to
Gal( ). Now the maximal abelian extension of in is the compositum of all cyclic
extensions of in . Since
acts trivially on each , we conclude that
is injective.
(3.1.8). 1 Acting on . For every every element can be uniquely expandedas
=
-
7/26/2019 Lectures on Local Fields
31/57
31
where is a prime element in .
Since : ur ur is continuous, it has exactly one extension : which acts as
.
We shall study the action of 1 on the multiplicative group.
The kernel of the homomorphism 1 is equal to and the
image is
; 1
=
for every
1.
Proof. For =
with
the condition ( ) = implies that (
) =
for
. Hence,
belongs to the residue field of and . Similarly one shows the
exactness of the sequence in the central term
+
.
Let
. We shall show the existence of a sequence
such that
1
mod
+1
and
+1 1
+1
.
Let =
0 with
, 0 1
. Let be such that
1 =
. Then 1 =
;
put 0 = .
Now assume that the elements 0 1
have already been constructed. Definethe element
+1 from the congruence
1
1
1 +
+1 +1 mod +2
There is an element
+1 such that
(
+1)
+1+
+1 =
+1
+1+
+1 0 mod
Now put
+1 =
(1 +
+1 +1). Then 1
1
+1
+2
and
+1 1
+1
.
There exists = lim
, and 1 = . When
the element can be
chosen in
as well.
Let be a finite Galois totally ramified extension. The extension ur ur is Galois with
the group isomorphic to that of . We may assume that the completion of ur is a subfield
of the completion of ur.
The extension is totally ramified of the same degree as . From (2.7) and (2.8) we
deduce that the extension is Galois with the group isomorphic to that of .
Let be such that 1 ( ). Then
belongs to the group
.
Proof. We have 1 = 1
for some
and Gal( ). By the pre-
vious proposition we have = 1
for some
. So ( 1 1
) 1 = 1 and
1 1
= with . Then
=
1
.
-
7/26/2019 Lectures on Local Fields
32/57
32
3.2. The Neukirch Map
(3.2.1). Let be a finite Galois extension of . Then ur = ur. Recall that Gal( ur )
consists of -powers of .
Put Frob( ) =
Gal( ur
) :
ur is a positive integer power of .
The setFrob( ) is closed with respect to multiplication;it is not closed with
respect to inversion and 1 Frob( ).
The fixed field of
Frob(
) is of finite degree over
, ur = ur, and
is the
Frobenius automorphism of .
Thus, the set Frob( ) consists of the Frobenius automorphisms of finite extensions
of in ur with Gal( ur
)
.
The map Frob( )
Gal(
)
is surjective.
Proof. The first assertion is obvious.
Since
ur
we deduce that
ur
ur
ur
. The Galois group of
ur
istopologically generated by
and isomorphic to , therefore it does not have nontrivial closed
subgroups of finite order. So the group Gal( ur ur) being a subgroup of the finite group
Gal( ur
ur) should be trivial. So ur = ur.
Put 0 = ur . This field is the fixed field of
ur =
, therefore 0 : = is
finite. We deduce that
: 0 =
ur :
ur
=
ur :
ur
=
: 0
is finite. Thus, is a finite extension.
Now
is a power of and ur = 0:
ur =
ur =
ur . Therefore,
=
. Certainly, the Frobenius automorphism of a finite extension of in ur with
Gal( ur ) belongs to Frob( ).
Denote by an extension in Gal( ur ) of . Let
Gal(
), then
0 is equal
to for some positive integer . Hence
1
acts trivially on 0, and so = belongs to Gal( 0). Let Gal(
ur
ur) be such that
= . Then for
= we
deduce that
ur =
and
=
=
. Then the element
Frob(
) is mapped
to
Gal(
).
(3.2.2).
Let be a finite Galois extension. Introduce
: Frob( )
mod
where is the fixed field of
Frob(
) and is any prime element of .
The map
is well defined. If
= id
then
(
) = 1.
Proof. Let
1
2 be prime elements in . Then
1 =
2
for a unit
. Let
be thecompositum of and . Since ur , the extension is unramified. From (3.1.5)
we know that =
for some . Hence
1 =
( 2 ) =
2
(
) =
2
(
)
-
7/26/2019 Lectures on Local Fields
33/57
33
We obtain that
1
2 mod
.
If
= id
then and therefore
.
(3.2.3). The definition of the Neukirch map is very natural from the point of view of thewell known principle that a prime element in an unramified extension should correspond to
the Frobenius automorphism (see Theorem (3.2.4) below) and the functorial property of the
reciprocity map (see (3.2.5) and (3.3.4)) which forces the reciprocity map
to be defined
as it is.
Already at this stage one can prove that the map
: Frob( )
induces the Neukirch homomorphism
: Gal( )
In other words,
(
) does not depend on the choice of
Frob( ) which extends
Gal( ), and moreover,
( 1)
( 2) =
( 1 2).
We will choose a different route, which is a little longer but perhaps is more satisfying.
The plan is the following: first we easily show the existence of
for unramifiedextensions and even prove that it is an isomorphism. Then we deduce some functorial properties
of
. To treat the case of totally ramified extensions in the next section, we introduce the
Hazewinkel homomorphism
which acts in the opposite direction to
. Calculating
composites of the latter with
we shall deduce the existence of
which is expressed
by the commutative diagram
Frob( )
id
Gal( )
Then using
we prove that
is a homomorphism and that its abelian part
ab
: Gal( )ab
is an isomorphism.
Then we treat the general case of abelian extensions and then Galois extensions reducing it
to the two cases described above and using functorial properties of
. This route not only
establishes the existence of
, but also implies its isomorphism properties.
(3.2.4).